Sliding Window
Author: Benjamin Qi
Contributors: Darren Yao, Andrew Wang, Chuyang Wang, Martin Lin
Maintaining data over consecutive subarrays.
Sliding Window
From CPH:
A sliding window is a constant-size subarray that moves from left to right through the array.
For each position of the window, we want to compute some information.
Focus Problem – try your best to solve this problem before continuing!
Implementation
The most straightforward way to do this is to store an sorted set of integers representing the integers inside the window. If the window currently spans the range , we observe that moving the range forward to only removes and adds to the window. We can support these two operations and query for the minimum / maximum in the set in .
Time Complexity:
C++
vector<int> maxSlidingWindow(vector<int> &nums, int k) {multiset<int> s;vector<int> ret;for (int i = 0; i < k; i++) { s.insert(nums[i]); }for (int i = k; i < nums.size(); i++) {ret.push_back(*s.rbegin());s.erase(s.find(nums[i - k]));s.insert(nums[i]);}ret.push_back(*s.rbegin());return ret;}
Java
static TreeMap<Integer, Integer> multiset = new TreeMap<Integer, Integer>();static void add(int x) {if (multiset.containsKey(x)) {multiset.put(x, multiset.get(x) + 1);} else {multiset.put(x, 1);}}static void remove(int x) {multiset.put(x, multiset.get(x) - 1);
Problems
Status | Source | Problem Name | Difficulty | Tags | |
---|---|---|---|---|---|
CSES | Normal | Show TagsPrefix Sums, Sliding Window | |||
CSES | Normal | Show TagsSet, Sliding Window | |||
CSES | Hard | Show TagsSet, Sliding Window |
With Two Pointers
In general, it is not required for the subarray to have constant size as long as both the left and right endpoints of the subarray only move to the right.
Focus Problem – try your best to solve this problem before continuing!
Solution
C++
int n;set<int> s;int a[200000], ans;int main() {int r = -1;cin >> n;F0R(i, n) cin >> a[i];F0R(i, n) {while (r < n - 1 && !s.count(a[r + 1])) s.insert(a[++r]);ans = max(ans, r - i + 1);s.erase(a[i]);}cout << ans;}
Python
n = int(input().strip())nums = [int(v) for v in input().split(" ")]ans = -float("inf")left, right = 0, 0unique_songs = set()while right < n:# Notice that all the songs in unique_songs are unique in each iteration.# We keep this property by shrinking the window before inserting nums[right].
Java
public class test {public static void main(String[] args) throws Exception {BufferedReader br =new BufferedReader(new InputStreamReader(System.in));int n = Integer.parseInt(br.readLine());StringTokenizer st = new StringTokenizer(br.readLine());int a[] = new int[n];for (int i = 0; i < n; i++) a[i] = Integer.parseInt(st.nextToken());int r = -1;HashSet<Integer> s = new HashSet<Integer>();
Problems
Status | Source | Problem Name | Difficulty | Tags | |
---|---|---|---|---|---|
CF | Easy | ||||
Gold | Easy | Show TagsSet, Sliding Window | |||
AC | Easy | Show TagsSliding Window | |||
CSES | Easy | Show Tags2P, Sliding Window | |||
APIO | Normal | Show TagsMedian, Sliding Window | |||
Gold | Normal | Show TagsSliding Window | |||
Platinum | Normal | Show TagsSliding Window | |||
APIO | Hard | Show TagsDP, Sliding Window | |||
IOI | Hard | Show TagsBinary Search, DP, Sliding Window | |||
IOI | Hard | Show TagsDP, Sliding Window |
Sliding Window Minimum in
Focus Problem – try your best to solve this problem before continuing!
Resources
Resources | ||||
---|---|---|---|---|
cp-algo | multiple ways to solve this |
Method 1 - Deque
Method 2 from cp-algo.
Resources | ||||
---|---|---|---|---|
CPH |
C++
vector<int> maxSlidingWindow(vector<int> &nums, int k) {deque<int> d;vector<int> ret;for (int i = 0; i < k; i++) {while (!d.empty() && nums[i] > nums[d.back()]) { d.pop_back(); }d.push_back(i);}for (int i = k; i < nums.size(); i++) {ret.push_back(nums[d.front()]);if (!d.empty() && d.front() <= i - k) { d.pop_front(); }
Java
static ArrayList<Integer> maxSlidingWindow(int[] nums, int k) {ArrayList<Integer> ret = new ArrayList<Integer>();ArrayDeque<Integer> d = new ArrayDeque<Integer>();for (int i = 0; i < k; i++) {while (!d.isEmpty() && nums[i] > nums[d.peekLast()]) { d.pollLast(); }d.addLast(i);}for (int i = k; i < nums.length; i++) {ret.add(nums[d.peekFirst()]);if (!d.isEmpty() && d.peekFirst() <= i - k) { d.pollFirst(); }
Python
from typing import Listdef maxSlidingWindow(self, nums: List[int], k: int) -> list[int]:d = collections.deque()def enqueue(i: int) -> None:while d and nums[d[-1]] <= nums[i]:d.pop()d.append(i)
Method 2 - Two Stacks
Method 3 from cp-algo. Not as common but nice to know!
We use two stacks and to simulate our maximum queue. Every time we add an element to it, we push the element itself and the maximum in stack after adding this element to . To pop out the front element from our queue, we just pop the top element from . Since elements in are stored in the order of how we added them, i.e. the last added element is at the top of , we just have to pop all of them out and push them into the stack when is empty. After that, these elements in will be in reversed order, i.e. the first added element will be at the top of , and we can pop them out as from normal stacks to simulate the dequeue operation of our queue. To find the maximum among all elements in our queue, we just have to return the maximum of both stacks, which is stored in the top element when we added it.
Then, we can solve the problem by removing the first element and adding a new element to the queue to simulate our sliding window. As each operation of our queue takes time, and we add each of elements once, we get a time complexity of .
C++
struct MaxQueue {/*** For each pair<int, int> e, e.first is the value of the element and* e.second is the maximum among all elements in that stack under element e.*/stack<pair<int, int>> s1, s2;/*** Get the maximum element in the MaxQueue.* It is the maximum of both stacks which are stored in s.top().second.
Java
class Solution {public int[] maxSlidingWindow(int[] nums, int k) {MaxQueue q = new MaxQueue();int n = nums.length;int[] maxVals = new int[n - k + 1];// Fill the queue with elements from the first windowfor (int i = 0; i < k; i++) { q.enqueue(nums[i]); }maxVals[0] = q.query();
Problems
Status | Source | Problem Name | Difficulty | Tags | |
---|---|---|---|---|---|
YS | Hard | ||||
Baltic OI | Hard |
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